f(x) = begin{cases} frac{(2x^2 - ax + 1) - (a | vedclass

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(D) For $f(x)$ to be continuous at $x = -1$,we must have $\lim_{x 	o -1} f(x) = f(-1) = k$. The limit is $\lim_{x 	o -1} \frac{(2-a)x^2 - (a+3b)x -

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$a - 3$

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(D) For $f(x)$ to be continuous at $x = -1$,we must have $\lim_{x 	o -1} f(x) = f(-1) = k$. The limit is $\lim_{x 	o -1} \frac{(2-a)x^2 - (a+3b)x - 1}{x+1}$. For the limit to exist,the numerator must be $0$ at $x = -1$. Substituting $x = -1$: $(2-a)(-1)^2 - (a+3b)(-1) - 1 = 0 \implies 2 - a + a + 3b - 1 = 0 \implies 1 + 3b = 0 \implies b = -\frac{1}{3}$. Substituting $b = -\frac{1}{3}$ into the numerator: $(2-a)x^2 - (a - 1)x - 1$. This can be factored as $(x+1)((2-a)x - 1)$. Thus,$\lim_{x 	o -1} \frac{(x+1)((2-a)x - 1)}{x+1} = \lim_{x 	o -1} ((2-a)x - 1) = (2-a)(-1) - 1 = -2 + a - 1 = a - 3$. Since the function is continuous,$k = a - 3$.

$f(x) = \begin{cases} \frac{(2x^2 - ax + 1) - (ax^2 + 3bx + 2)}{x + 1} & ; x \neq -1 \\ k & ; x = -1 \end{cases}$ is a real-valued function. If $a, b, k \in R$ and $f$ is continuous on $R$,then $k =$

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